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I Function of 2 variables, max/min test, D=0 and linear dependence

  1. Dec 19, 2017 #1
    ##f(x,y)##

    a critical point is given by ##f_x=0## and ##f_y=0## simultaneously.

    the test is:

    ##D=f_{xx}f_{yy}-(f_{xy})^2 ##

    if ##D >0 ## and ##f_{xx} <0 ## it is a max
    if ##D >0 ## and ##f_{xx} >0 ## it is a min
    ##D >0 ## is is a saddle
    if ##D =0 ## it is inconclusive, and ##f_x## and ##f_y## are not linear independent.

    I'm stuck on the ##D=0## comment re linear independence. So is this saying that ##x## and ##y## are not linear indepedent?

    So if i take an arbitary function ##f(x,y) ## and ##y=h(x)##, h some linear function, then I should get ##D=0## or not?
     
  2. jcsd
  3. Dec 19, 2017 #2

    Mark44

    Staff: Mentor

    It is saying that the two first partials are linearly dependent, not that x and y are linearly dependent.
     
  4. Dec 19, 2017 #3
    What is characteristic of a function whose partials are linearly dependent?
     
  5. Dec 19, 2017 #4

    Mark44

    Staff: Mentor

    If two functions (partials in this case) are linearly dependent, then each will be a constant multiple of the other.
     
  6. Dec 19, 2017 #5
    yup, so if i have a ##f(x,y)## function all term must be a multiple of ##(xy)^n## , ##n \in R## or if it contains terms solely of ##x## or ##y## they must be of at most the power of ##2##? (To conclude the partials ##f_x## and ##f_y## are linear dependent, not higher partials)?

    edit: re second comment is incorrect, but would yield ##D=0## without implying that ##f_x## and ##f_y## are linearly independent? so in OP comment only holds for higher powers? or a function without sole terrms of x or y if the power is lower
     
  7. Dec 19, 2017 #6

    Mark44

    Staff: Mentor

    I haven't worked out an example. You want a function f(x, y) such that fx = fy = 0, and for which the discriminant D = 0.
     
  8. Dec 19, 2017 #7
    apologies edited post since above
     
  9. Dec 19, 2017 #8

    StoneTemplePython

    User Avatar
    Science Advisor
    Gold Member

    this is a typo and should say less than zero.



    This is a lot easier in my view if you know bits about linear algebra. Do you know what a Hessian is? How about a quadratic form?

    consider the 2 variable function (I used ##x_1, x_2## instead of ##x,y##:

    edit: (due to some domain subtleties, the below is a better example than the prior function I mentioned)

    ##f(x_1,x_2) = (x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1 x_2##

    The Hessian is given by
    ##
    \mathbf H = \begin{bmatrix}
    2 & 2\\
    2 & 2
    \end{bmatrix} = 2 \big(\mathbf 1 \mathbf 1^T\big)##

    This is a rank one matrix and hence has determinant of zero. (Equivalently compute the determinant directly and see ##2*2 - 2*2 = 0## )
     
    Last edited: Dec 19, 2017
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